BYJU'S 2024 papers
Previous year BYJU’S placement papers with questions and solutions
This page is a working set of BYJU’S placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what BYJU’S actually asked in the latest cycle, how hard the rounds were, and which themes (DSA, system design, aptitude, or role-specific topics) mattered most. Practice the problems below under timed conditions, then cross-check with the interview and preparation guides if you are targeting an upcoming BYJU’S drive.
| Section | Questions | Time | Difficulty |
|---|---|---|---|
| Coding Problems | 2-3 | 90 min | Medium-Hard |
This section contains practice questions styled on BYJU’S placement papers 2025 (recent-cycle pattern), with worked solutions. Use them as timed sectional drills - from student reports drives vary by college and role, so treat this as a high-signal practice bank, not an official paper dump.
Problem Statement: Return the level-order traversal of a binary tree.
Example:
Input: [3,9,20,null,null,15,7]Output: [[3],[9,20],[15,7]]Solution (Java):
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) return res; Queue<TreeNode> q = new ArrayDeque<>(); q.add(root); while (!q.isEmpty()) { int sz = q.size(); List<Integer> level = new ArrayList<>(); for (int i = 0; i < sz; i++) { TreeNode n = q.poll(); level.add(n.val); if (n.left != null) q.add(n.left); if (n.right != null) q.add(n.right); } res.add(level); } return res;}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Find the contiguous subarray with the largest sum.
Example:
Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4]Output: 6 // [4, -1, 2, 1]Solution (Java):
public int maxSubArray(int[] nums) { int best = nums[0], cur = nums[0]; for (int i = 1; i < nums.length; i++) { cur = Math.max(nums[i], cur + nums[i]); best = Math.max(best, cur); } return best;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Return true if the string reads the same forward and backward (ignore case).
Example:
Input: "Level"Output: trueSolution (Java):
public boolean isPalindrome(String s) { s = s.toLowerCase(); int i = 0, j = s.length() - 1; while (i < j) { if (s.charAt(i++) != s.charAt(j--)) return false; } return true;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Return true if the linked list has a cycle.
Example:
Input: 3→2→0→-4→(back to 2)Output: trueSolution (Java):
public boolean hasCycle(ListNode head) { ListNode slow = head, fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; } return false;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given a string, return it reversed.
Example:
Input: "placement"Output: "tnemecalp"Solution (Java):
public String reverse(String s) { return new StringBuilder(s).reverse().toString();}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Given an array of integers and a target, return indices of two numbers that add up to target.
Example:
Input: nums = [2, 7, 11, 15], target = 9Output: [0, 1]Solution (Java):
public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int need = target - nums[i]; if (map.containsKey(need)) return new int[]{map.get(need), i}; map.put(nums[i], i); } return new int[]{};}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Return the first non-repeating character in a string, or ‘_’ if none.
Example:
Input: "swiss"Output: 'w'Solution (Java):
public char firstUnique(String s) { int[] freq = new int[256]; for (char c : s.toCharArray()) freq[c]++; for (char c : s.toCharArray()) if (freq[c] == 1) return c; return '_';}Time Complexity: O(n)
Space Complexity: O(1)
Solution:
HAVING filters aggregates; WHERE filters rows before grouping.
Answer: HAVING
Solution:
Unbalanced partitions (already sorted with bad pivot) → O(n²).
Answer: O(n²)
Solution:
This is the definition of Encapsulation (often paired with abstraction in interviews).
Answer: Encapsulation
Solution:
OS uses demand paging (and sometimes segmentation) to implement virtual memory.
Answer: Demand paging
Based on recent candidate experiences from 2025 BYJU’S interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
Success Tips:
For detailed interview experiences from 2025, visit BYJU’S Interview Experience page.
BYJU'S 2024 papers
Previous year BYJU’S placement papers with questions and solutions
BYJU'S coding questions
Complete collection of BYJU’S coding problems with solutions
BYJU'S interview experience
Real interview experiences from successful candidates
BYJU'S preparation guide
Comprehensive preparation strategy for BYJU’S placement
BYJU'S main page
Complete BYJU’S placement guide with eligibility, process, and salary
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